A six coordination complex of formula `CrCl_(3)*6H_(2)O` has green colour. A 0.1 M solution of the complex when treated with excess of `AgNO_(3)` gas 28.7g of white precipitate. The formula of the complex would be:

Options

A. `[Cr(H_(2)O)_(6)]Cl_(3)`
B. `[CrCl(H_(2)O)_(5)]Cl_(2)*H_(2)O`
C. `[CrCl_(2)(H_(2)O)_(4)]Cl*2H_(2)O`
D. `[Cr(H_(2)O)_(3)Cl_(3)]`

Answer

To determine the formula of the coordination complex `CrCl3·6H2O`, we will follow these steps:

### Step 1: Understand the Reaction
When the complex `CrCl3·6H2O` is treated with excess `AgNO3`, the chloride ions (Cl⁻) present in the solution will react with silver nitrate to form silver chloride (AgCl), which is a white precipitate.

### Step 2: Calculate the Moles of AgCl
Given that 28.7 grams of AgCl precipitate is formed, we need to calculate the number of moles of AgCl produced. The molar mass of AgCl is approximately 143.32 g/mol.

\[ \text{Moles of AgCl} = \frac{\text{mass}}{\text{molar mass}} = \frac{28.7 \, \text{g}}{143.32 \, \text{g/mol}} \approx 0.2 \, \text{moles} \]

### Step 3: Relate Moles of AgCl to Moles of Cl⁻
Each mole of AgCl corresponds to one mole of Cl⁻. Therefore, if 0.2 moles of AgCl are produced, it means that 0.2 moles of Cl⁻ ions were present in the solution.

### Step 4: Determine the Number of Chloride Ions
Since the complex is `CrCl3·6H2O`, it contains 3 chloride ions in its coordination sphere. However, the formation of 0.2 moles of Cl⁻ indicates that there are additional chloride ions outside the coordination sphere.

### Step 5: Calculate the Total Chloride Ions
From the reaction, we can conclude that:
- 3 Cl⁻ ions are part of the coordination sphere.
- The remaining Cl⁻ ions must be from the excess chloride ions in solution.

Since we have 0.2 moles of Cl⁻ and we know 3 of them are part of the complex, we can deduce that:
\[ \text{Total Cl⁻} = 0.2 \, \text{moles} \]
This indicates that there are 2 Cl⁻ ions outside the coordination sphere.

### Step 6: Write the Final Formula
The final formula of the complex can be written as:
\[ \text{CrCl}_3 \cdot 6\text{H}_2\text{O} \]
This indicates that the complex contains 3 chloride ions coordinated to the chromium ion and 2 chloride ions remaining outside the coordination sphere.

### Conclusion
Thus, the formula of the complex is `CrCl3·6H2O`.